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Math Help - Challenging Integral

  1. #1
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    Challenging Integral

    solve  \int \frac{dx}{ sin^5{x} + cos^5{x} }


    and


     \int \frac{dx}{ sin^7{x} + cos^7{x} }
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  2. #2
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    Quote Originally Posted by simplependulum View Post

    solve  \int \frac{dx}{ sin^5{x} + cos^5{x} } and  \int \frac{dx}{ sin^7{x} + cos^7{x} }
    i'll solve the first one only and i'll leave the second integral to you or whoever likes trouble! first we need the following magical identity:

    \frac{1}{\sin^5 x + \cos^5 x} = \frac{4}{5} \left[\frac{1}{\sin x + \cos x} + \frac{(\sin x + \cos x)((\sin x - \cos x)^2 - 2)}{(\sin x - \cos x)^4 - 4(\sin x - \cos x)^2 - 1} \right].

    now finding \int \frac{dx}{\sin x + \cos x} is straightforward. to find I= \int \frac{(\sin x + \cos x)((\sin x - \cos x)^2 - 2)}{(\sin x - \cos x)^4 - 4(\sin x - \cos x)^2 - 1} \ dx substitute: \sin x - \cos x = t. then (\sin x + \cos x) \ dx = dt and thus

    I=\int \frac{t^2-2}{t^4 - 4t^2 - 1} \ dt. finally let a=2+\sqrt{5} and b=2-\sqrt{5} to get: I=\frac{1}{2} \left[\int \frac{dt}{t^2-a} + \int \frac{dt}{t^2 - b} \right]. note that b < 0. now everything is nice and easy and you're good to go!



    Remark: i think this is true that for any n \in \mathbb{N} we have \frac{1}{\sin^{2n+1}x + \cos^{2n+1} x} = \frac{2n}{2n+1}\left[\frac{1}{\sin x + \cos x} + (\sin x + \cos x)R((\sin x - \cos x)^2) \right], where R(z) is some rational function of z.

    actually i think R(z)=\frac{p(z)}{q(z)}, where both p(z) and q(z) are monic polynomials with integer coefficients! for example, in your problem: p(z)=z-2, \ q(z)=z^2-4z-1. i also think that we

    always have \deg q(z) = 1+\deg p(z). these claims are either wrong or easy to prove! ... let's just leave it as a challenge for whoever has time and patience to try it.

    by the way, the identity \frac{1}{\sin^3x + \cos^3x}=\frac{2}{3} \left[\frac{1}{\sin x + \cos x} + \frac{\sin x + \cos x}{(\sin x - \cos x)^2 + 1} \right] shows that my claims are not based on nothing!
    Last edited by NonCommAlg; June 22nd 2009 at 08:27 AM. Reason: nothing important!
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  3. #3
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    My solution is to substitute  t = \sin{x}+\cos{x}
    so,
     dt = (\cos{x} - \sin{x} )dx

     \sin{x} \cos{x} = \frac{t^2 - 1}{2}

     dx = \frac{dx}{ \sqrt{ 2 - t^2 } }

    Although it looks very troublesome , it become quite well after separating it into fractions .

    However , i think your method is better than what i am using .
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